Sketch, on the same diagram, the graphs of $y = |2x + 3|$ and $y = -2x^2 - 8x - 3$.
(You are not required to label any axial intercepts and stationary points.) Solve exactly the inequality $|2x + 3| > -2x^2 - 8x - 3$.

1. Determine the shape and vertex of $y = |2x + 3|$.

The graph of $y = |2x + 3|$ is a V-shaped curve opening upwards.
Its vertex (minimum point) occurs where the expression inside the modulus is zero:
$2x + 3 = 0 \implies x = -1.5$
Thus, the vertex is at $(-1.5, 0)$.

2. Determine the shape and vertex of $y = -2x^2 - 8x - 3$.

The graph of $y = -2x^2 - 8x - 3$ is a parabola opening downwards (since the coefficient of $x^2$ is $-2$, which is negative).
The $x$-coordinate of its maximum point is given by $x = -\frac{b}{2a}$:
$x = -\frac{-8}{2(-2)} = -2$
The corresponding $y$-coordinate is:
$y = -2(-2)^2 - 8(-2) - 3 = -8 + 16 - 3 = 5$
Thus, the vertex of the parabola is at $(-2, 5)$.

3. Determine the points of intersection to establish relative positions.

Set the equations equal to find the intersections:
$|2x + 3| = -2x^2 - 8x - 3$

Case 1 ($x \ge -1.5$):
$2x + 3 = -2x^2 - 8x - 3$
$2x^2 + 10x + 6 = 0 \implies x^2 + 5x + 3 = 0$
$x = \frac{-5 \pm \sqrt{25 - 12}}{2} = \frac{-5 \pm \sqrt{13}}{2}$
Since $x \ge -1.5$, we accept $x = \frac{-5 + \sqrt{13}}{2} \approx -0.70$.

Case 2 ($x < -1.5$):
$-(2x + 3) = -2x^2 - 8x - 3$
$2x^2 + 6x = 0 \implies 2x(x + 3) = 0$
Since $x < -1.5$, we accept $x = -3$.

The graphs intersect at two distinct points: $x = -3$ and $x \approx -0.70$.

4. Sketch both graphs on the same axes.

Draw the coordinate axes.

• Sketch the V-shaped graph of $y = |2x + 3|$ touching the negative $x$-axis.
• Sketch the inverted parabola $y = -2x^2 - 8x - 3$ with its maximum point to the left of the V-shape's vertex.
• Ensure the parabola crosses the V-shaped graph at exactly two points.
  (As per the instructions, no labels for axial intercepts or stationary points are required.)

1. Find the intersection points by considering the positive case of the modulus.

To find where the graphs intersect, we solve $|2x + 3| = -2x^2 - 8x - 3$.

For $2x + 3 \ge 0$ (i.e., $x \ge -1.5$):
$2x + 3 = -2x^2 - 8x - 3$
$2x^2 + 10x + 6 = 0$
$x^2 + 5x + 3 = 0$
Using the quadratic formula:
$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(3)}}{2} = \frac{-5 \pm \sqrt{13}}{2}$
Since $x \ge -1.5$, we reject $x = \frac{-5 - \sqrt{13}}{2}$ (which is $\approx -4.3$), leaving $x = \frac{-5 + \sqrt{13}}{2}$.

2. Find the intersection points by considering the negative case of the modulus.

For $2x + 3 < 0$ (i.e., $x < -1.5$):
$-(2x + 3) = -2x^2 - 8x - 3$
$-2x - 3 = -2x^2 - 8x - 3$
$2x^2 + 6x = 0$
$2x(x + 3) = 0$
$x = 0 \text{ or } x = -3$
Since $x < -1.5$, we reject $x = 0$, leaving $x = -3$.

3. Use the graphical representation to determine the final solution intervals.

The inequality $|2x + 3| > -2x^2 - 8x - 3$ requires the graph of $y = |2x + 3|$ to be above the graph of $y = -2x^2 - 8x - 3$.

Referring to the sketch from part (i), the V-shaped modulus graph opens upwards, and the quadratic graph is an inverted parabola. The V-shape lies strictly above the parabola for $x$-values outside the interval bounded by the intersection points.

Therefore, the solution to the inequality is:
$x < -3 \text{ or } x > \frac{-5 + \sqrt{13}}{2}$